Principle used

In YDSE, if a thin transparent slab of thickness $t$ and refractive index $\mu$ is introduced in one path, the optical path increases by $\Delta = (\mu-1)t$

Here, slabs are introduced in both paths, so the net additional optical path difference is the difference of their optical path increases: $\Delta_{\text{net}} = (\mu_2-1)t-(\mu_1-1)t = (\mu_2-\mu_1)t$

Given data

$t=0.1\text{ mm}$ $\mu_1=1.51,\quad \mu_2=1.55$ $\lambda = 4000\,\text{\AA} = 4000\times 10^{-10}\text{ m} = 4\times 10^{-7}\text{ m}$

Calculate net path difference

$\Delta_{\text{net}} = (1.55-1.51)(0.1\text{ mm})$ $=0.04\times 0.1\text{ mm}$ $=0.004\text{ mm}$

Converting to meters: $0.004\text{ mm} = 4\times 10^{-6}\text{ m}$

Shift in number of fringes

Number of fringes shifted is $n=\frac{\Delta_{\text{net}}}{\lambda}$

$n=\frac{4\times 10^{-6}}{4\times 10^{-7}}=10$

Conclusion

The central bright fringe shifts by $\boxed{10}$ fringes.