Let the magnetic field of the $200$ turns be $B_1$. The formula of magnetic field is given by $B=\frac{{\mu }_{0}i}{2r}$

Let $r=1000 \mathrm{cm}$. Thus, 

$B_1=\frac{n{\mu }_0\times I}{2\times 10}=\frac{200{\mu }_{0}I}{2\times 10}$

The flux through the coil of radius $r_1=1 \mathrm{cm}$ is $\varphi =10\times \left(\overrightarrow{B}·\overrightarrow{A}\right)$$=10\times B_1\pi {r^2}_1$

$\Rightarrow \varphi =10\times \frac{200{\mu }_{0}I}{2\times 10}\times \pi {\left(0.01\right)}^2$ 

$$\begin{gathered} L=10\times \frac{200{\mu }_0}{2\times 10}\times \pi {\left(0.01\right)}^2 \\ =10\times \frac{200\left(4\pi \times {10}^{-7}\right)}{2\times 10}\times \pi {\left(0.01\right)}^2 \\ =4\times {10}^{-8} \mathrm{H} \end{gathered}$$