$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{k{x^2}}&{if}&{x \le 2}\\{3,}&{if}&{x > 2}\end{array}} \right.$ at $x = 2.$

We have,$f(2) = 4k$

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} 3 = 3$

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} {(2 - h)^2}k = 4k$

For continuity at x $=$ 2, we have

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2)$

$\Rightarrow$ $4k = 3 \Rightarrow k = \cfrac{3}{4}$