I believe that the answer is that there is no such isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of my argument:

First, take $\mathbb{H}\subset\mathbb{C}$ to be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ is then given by $|z|\ge 1$ and $|x|\le \tfrac12$, where one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$.

Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z_i$. There will be a hyperbolic disk $D_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z_i$, and because the curvature is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.

On the other hand, let $\epsilon>0$ be a very small positive number and consider the part $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles:

1. $C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;

2. $C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and

3. $C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.

In particular, since $\epsilon>0$ is taken to be quite small, each of these curves has length less than, say, $4\epsilon$.

The images of each of these three closed curves under $f$ must therefore lie in an Euclidean ball of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls of radius $4\epsilon$ found above and look at the $4\epsilon$-neighborhood of that plane.)

Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.

However, if we take $\epsilon<R/4$ sufficiently small, the disk $D_r(z)$ will lie entirely within $M_\epsilon$ and hence the $f$-image of $D_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<R$, which is obviously impossible.

Thus, $f$ cannot exist.

There is no isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of an argument:

First, let $\mathbb{H}\subset\mathbb{C}$ be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ on $\mathbb{H}$ is then defined by the inequalities $|z|\ge 1$ and $|x|\le \tfrac12$. Then one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$.

Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z_i$. There will be a hyperbolic disk $D_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z_i$. Because the Gauss curvature of $X(1)$ is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.

Meanwhile, let $\epsilon>0$ be a very small positive number and consider the subset $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles:

1. $C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;

2. $C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and

3. $C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.

In particular, when $\epsilon>0$ is taken to be sufficiently small, each of these curves has total length less than $4\epsilon$.

Thus, each $f(C_i)$ must therefore lie in an Euclidean ball $B_i$ of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls $B_i$ of radius $4\epsilon$ and look at the $4\epsilon$-neighborhood of that plane.)

Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.

However, if we take $\epsilon<R/4$ sufficiently small, the disk $D_r(z)$ will lie entirely within $M_\epsilon$ and hence the $f$-image of $D_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<R$, which is obviously impossible.

Thus, such an $f$ cannot exist.

I believe that the answer is that there is no such isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of my argument:

First, take $\mathbb{H}\subset\mathbb{C}$ to be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ is then given by $|z|\ge 1$ and $|x|\le \tfrac12$, where one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$.

Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ other than $z_i$. There will be a disk $D_r(z)$ of some radius $r>0$ about $z$ that is immersed isometrically, and because the curvature is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.

On the other hand, let $\epsilon>0$ be a very small positive number and consider the part $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles:

1. $C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;

2. $C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and

3. $C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.

In particular, since $\epsilon>0$ is taken to be quite small, each of these curves has length less than, say $4\epsilon$.

The images of each of these three closed curves under $f$ must therefore lie in an Euclidean ball of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$.

Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.

However, if we take $\epsilon<R/4$ sufficiently small, the disk $D_r(z)$ will lie entirely within $M_\epsilon$ and hence the $f$-image of $D_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<4$, which is obviously impossible.

Thus, $f$ cannot exist.

I believe that the answer is that there is no such isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of my argument:

First, take $\mathbb{H}\subset\mathbb{C}$ to be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ is then given by $|z|\ge 1$ and $|x|\le \tfrac12$, where one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$.

Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z_i$. There will be a hyperbolic disk $D_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z_i$, and because the curvature is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.

On the other hand, let $\epsilon>0$ be a very small positive number and consider the part $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles:

1. $C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;

2. $C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and

3. $C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.

In particular, since $\epsilon>0$ is taken to be quite small, each of these curves has length less than, say, $4\epsilon$.

The images of each of these three closed curves under $f$ must therefore lie in an Euclidean ball of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls of radius $4\epsilon$ found above and look at the $4\epsilon$-neighborhood of that plane.)

Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.

However, if we take $\epsilon<R/4$ sufficiently small, the disk $D_r(z)$ will lie entirely within $M_\epsilon$ and hence the $f$-image of $D_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<R$, which is obviously impossible.

Thus, $f$ cannot exist.