Solution:

The provided function is

$f(x)= \begin{cases}k x^{2}, & \text { if } x \leq 2 \\ 3, & \text { if } x>2\end{cases}$

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=3$

$\lim _{x \rightarrow 2^{-}} f(x)=3$ and $f(2)=3$

$k \times 2^{2}=3$

As it implies that, $k=\frac{3}{4}$

When the value of $k=3/4$, as a result $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} \frac{3}{4}(2-h)^{2}=3$

As a result, $f(x)$ is continuous at $x=2$ when $k=\frac{3}{4}$.