Answer
The change in the film thickness is $1.89\mu\mathrm{m}$.
Work Step by Step
For a bright fringe to form on the narrow end of the film, the wave that transmits straight through (undergoing refraction twice and no phase change) must be in phase with the wave that bounces twice within the film before transmitting (and hence has undergone phase change partly due to reflection and partly due to the path difference).
That is, $2d_1=\left(m_1+{1\over 2}\right)\lambda_n$
Where, $d_1$ is the thickness of the film on the narrow end.
Likewise it is with the opposite broader end (of thickness $d_2$). For the 10th bright fringe to be formed there,
$2d_2=\left(m_1+9+{1\over 2}\right)\lambda_n$
Subtracting the two equations, we get,
\begin{align*}
2(d_2-d_1)&=9\lambda_n\\
d_2-d_1&={9\lambda\over 2n}\\
&={9\times 630 \times 10^{-9}\over 2\times 1.50}\\
&=1890\times 10^{-9}\\
&=1.89\mu\mathrm{m}
\end{align*}
Therefore, the change in the film thickness is $1.89\mu\mathrm{m}$.
