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A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.

Added: The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a representation $Q\to GL(4,2)$ is given by $$x \mapsto \begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad y \mapsto \begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$ Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$.

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.

Added: The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a faithful representation $Q\hookrightarrow GL(4,2)$ is given by $$x \mapsto \begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad y \mapsto \begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$ Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$.

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.

Added: The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a representation $Q\to GL(4,2)$ is given by $$x \mapsto \begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad y \mapsto \begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$ Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$.

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $\mathfrak{L}$ resp. $\mathfrak{R}$ be the set of left resp. right ideals of $\mathbb{F}_2[Q]$. By the above, we have an injective map $\mathfrak{L} \to \mathfrak{R},\;I \mapsto I$ between finite sets. Hence it suffices to show that $\mathfrak{L}, \mathfrak{R}$ have the same cardinality.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal. Hence $i$ induces maps $f: \mathfrak{L} \to \mathfrak{R}, I \to i(I)$, $g: \mathfrak{R} \to \mathfrak{L}, I \to i(I)$ which are inverse to each other since $i^2 = \text{id}$.

A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8.

For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper.

Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal.

Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.