Uniform but time-varying magnetic field  $B=\frac{B_{0}t}{T}$ is present in the cylindrical region of radius $R$. There is a circular conducting ring of radius $a$ is kept at distance $5a$. (Center to center distance)

Due to changing magnetic field, induced electric field is set up.

Since the magnetic field (inward) is increasing with time, the direction of induced electric field will be anticlockwise.

The value of induced emf in ring from B to C is not zero. It is because the angle between direction of induced electric field and length of differential element is acute at all points along the arc $BC$.    Option (a) is wrong.

Induced electric field at every point along the diameter $BD$ of the ring is perpendicular to the diameter $BD$ . Hence, $e={\int }_B^D\overrightarrow{E}·\overrightarrow{dl}=0$  Option (b) is correct.

Consider an arc of a circle with centre at centre of magnetic field (say $O$ ) and radius $x$  such that it intersects the conducting ring at two points, say $P$  and $Q$. Let centre of conducting ring be $O\text{'}$. Let angle between lines $OO\text{'}$ and $OP$ be $\theta$.

Induced electric field along the arc $PQ$ is $E$ given by

 $E\left(2\pi x\right)=\pi R^2\left(\frac{dB}{dt}\right)$

$E=\frac{R^2}{2x}\left(\frac{dB}{dt}\right)=\frac{R^2{B}_0}{2xT}$

Induced emf between points $P$ and $Q$  is

$e=E\left(PQ\right)=\frac{R^2{B}_0}{2xT}\left(2\theta x\right)=\frac{R^2{B}_0\theta }{T}\propto \theta$

Maximum value of $\theta$ is when the line $OP$ becomes a tangent to the ring. Then

$\sin \theta =\frac{a}{5a}=\frac{1}{5}$

$e_{\max }=\frac{R^2{B}_0}{T}{\sin }^{-1}\left(\frac{1}{5}\right)$     Option (c) is correct.

The value of induced emf in ring from A to C is not zero. It is because the angle between direction of induced electric field and length of differential element is acute at all points along the line $AC$.    Option (d) is wrong.