Step 1:
Let $O$ be the centre of the inscribed circle of triangle ABC.We have drawn another circle passing through $O,B$ and $C$.Suppose radius is equal to $R$.
Applying sine rule in $\Delta OBC$,we get
$\large\frac{a}{\sin \angle BOC}$$=2R$
$\Rightarrow R=\large\frac{a}{2\sin \angle BOC}$-----(1)
Now,since $O$ is the centre of the inscribed circle,hence $BO$ and $CO$ are bisections of angles $B$ and $C$ respectively.
$\angle OBC=\large\frac{B}{2}$
$\angle OCB=\large\frac{C}{2}$
$\angle BOC=180^{\large\circ}-\big(\large\frac{B}{2}+\frac{C}{2}\big)$
$\qquad\;\;=180^{\large\circ}-\big(90^{\large\circ}-\large\frac{A}{2}\big)$
$\qquad\;\;=90^{\large\circ}+\large\frac{A}{2}$
Step 2:
Substituting the value in (1) we get,
$R=\large\frac{a}{2\sin (90^{\large\circ}+\large\frac{A}{2}\big)}$
$\;\;=\large\frac{a}{2}$$\sec\large\frac{A}{2}$
$\;\;=\large\frac{a}{R}$$\sec\large\frac{A}{2}$
$\;\;=2$
Hence (c) is the correct answer.