Here are some details on how to compute the rank of $\pi_3$. A K3 surface $X$ is, in addition to being simply connected, a compact Kähler manifold, and such spaces are known to be formal in the sense of rational homotopy theory; this means that their rational homotopy can be computed by finding a Sullivan minimal model of their rational cohomology rings, and in particular only depends on the rational cohomology ring. (Terzić's paper linked to in Reimundo's answer uses instead that a compact oriented simply connected $4$-manifold is formal.)

Here is, briefly, how this computation works out, at least if I'm not misreading something. The goal is to build a graded rational vector space $V^{\bullet} = \bigoplus_{k \ge 2} V^k$ and a differential $d$ on the exterior algebra $\Lambda^{\bullet}(V)$ such that

- the cohomology of $(\Lambda^{\bullet}(V), d)$ agrees with $H^{\bullet}(X, \mathbb{Q})$,
- $dV$ is contained in $\Lambda^{\ge 2}(V)$.

The machinery of rational homotopy theory, together with the fact that $X$ is formal and has homology of finite type, then guarantees that we have a natural identification

$$\pi_{\bullet}(X) \otimes \mathbb{Q} \cong \text{Hom}_{\mathbb{Q}}(V^{\bullet}, \mathbb{Q}).$$

In particular, $\dim \pi_{\bullet}(X) \otimes \mathbb{Q} = \dim V^{\bullet}$. So to compute $\dim \pi_3(X) \otimes \mathbb{Q}$ it suffices to figure out how many elements we need in $V^3$.

We already know that we need $\dim V^2 = b_2$, where $b_2 = \dim H^2(X, \mathbb{Q}) = 22$. The cup product $H^2(X, \mathbb{Q}) \times H^2(X, \mathbb{Q}) \to H^4(X, \mathbb{Q})$ takes the form

$$\alpha \cup \beta = Q(\alpha, \beta) \gamma$$

where $Q(\alpha, \beta)$ is the intersection form and $\gamma$ is a generator of $H^4(X, \mathbb{Q})$. The only way to impose these relations on the cohomology of $(\Lambda^{\bullet}(V), d)$ is to introduce elements in $V^3$ whose differentials will impose those relations. Explicitly, let $e_1, e_2, \dots e_{22}$ be an orthogonal basis for $H^2(X, \mathbb{Q})$ with respect to the intersection form, so that $Q(e_i, e_j)$ is some nonzero multiple of $\delta_{ij}$. For $i \neq j$ we need to introduce $\frac{b_2(b_2 - 1)}{2} = 231$ new elements of $V^3$, call them $f_{ij}, i \neq j$, so that we can impose the relations

$$d f_{ij} = e_i \cup e_j.$$

For $i = j$ we need to introduce introduce $b_2 - 1 = 21$ new elements of $V^3$, call them $f_i, 1 \le i \le 21$, so that we can impose the relations

$$d f_i = \frac{e_i \cup e_i}{Q(e_i, e_i)} - \frac{e_{i+1} \cup e_{i+1}}{Q(e_{i+1}, e_{i+1})}.$$

(We cannot introduce the generator of $H^4(X, \mathbb{Q})$ into $V^4$ because we cannot impose a relation that is linear in this generator, so instead we impose the relation that all of the $e_i$ square, up to a normalization, to the same thing.)

Altogether, we get

$$\dim V^3 = {b_2 \choose 2} + (b_2 - 1) = {b_2 + 1 \choose 2} - 1 = 252$$

as expected from the other answers.