If $\delta$ is the minimum degree, given as $3.$
Take a vertex of degree $3.$ This vertex has $3$ edges incident to it.
These $3$ edges further lead to $3$ more vertices. Now these vertices can have a minimum degree of $3.$
We get the following planar graph$.$
This is the planar graph with minimum degree $3$ for each vertex.
From this graph, we can say that $3n\leq 2e \qquad \to(1)$
As per Euler's formla $: n − e + f = 2 \implies e=n+f-2 \qquad \to (2)$
From $(1)$ and $(2)$
$ n+f-2\geq \frac{3n}{2}$
$\implies f\geq \frac{3n}{2} - n + 2$
$\implies f\geq \frac{n}{2} + 2 $ $\text{(No of faces is at-least $(\frac{n}{2} + 2)$ )}$