If $f(x)=\{\begin{array}{ll}K{x}^2& \text{if }x\le 2\\ 3& \text{if }x>2\end{array}$ is continuous at $x=2$, then the value of $K$ is:

For a function to be continuous at $x=a$, the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function must be equal.LHL at $x=2$:
$$\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2}{\lim }K{x}^2=K(2{)}^2=4K$$
RHL at $x=2$:
$$\underset{x\to 2^{+}}{\lim }f(x)=3$$
Since the function is continuous at $x=2$:
$$4K=3⟹K=\frac{3}{4}$$