Answer
$1.5$
Work Step by Step
We know that the number of shifted fringes is given by
$$\Delta m=m_2-m_1$$
where $m_1=2d/\lambda_{\rm vaccum}$, and $m_2=2d/\lambda_{\rm glass}$ where $\lambda_{\rm glass}=\lambda_{\rm vaccum}/n_{\rm glass}$
So,
$$\Delta m= \dfrac{2dn_{\rm glass}}{\lambda_{\rm vaccum}}- \dfrac{2d }{\lambda_{\rm vaccum}}$$
$$\Delta m= \dfrac{2d(n_{\rm glass}-1)}{\lambda_{\rm vaccum}} $$
Solving for $n_{\rm glass}$;
$$n_{\rm glass}=\dfrac{\lambda_{\rm vaccum}\Delta m}{2d}+1$$
Plugging the known;
$$n_{\rm glass}=\dfrac{(500\times 10^{-9})(200)}{2(0.1\times 10^{-3})}+1=\color{red}{\bf 1.50}$$
