Answer
${\bf 400}\;\rm turn$
Work Step by Step
We know that the change in the current will create an induced emf.
And we know that the inductance is given by
$$\Delta V_L=L_{\rm solenoid}\dfrac{dI}{dt}$$
$$L_{\rm solenoid}=\dfrac{\Delta V_L}{ dI/dt}\tag 1$$
We are given the flux through one turn of the solenoid, so
$$\dfrac{\Phi}{N}=\dfrac{B_{\rm solenoid}A}{N}=\dfrac{\mu_0NI}{l}\dfrac{A}{N}$$
$$\dfrac{\Phi}{N} =\left[\dfrac{\mu_0A}{l}\right]I\tag 2$$
Recalling that
$$L_{\rm solenoid}=\dfrac{\mu_0 N^2A}{l}$$
Rearranging,
$$\dfrac{L_{\rm solenoid}}{N^2}=\dfrac{\mu_0 A}{l} $$
Plug into (2),
$$\dfrac{\Phi}{N} =\left[\dfrac{L_{\rm solenoid}}{N^2}\right]I $$
Solving for $N$;
$$ N=\left[\dfrac{L_{\rm solenoid}}{\Phi}\right]I $$
Plug from (1);
$$ N= \dfrac{\dfrac{\Delta V_L}{ dI/dt} }{\Phi} I$$
Plug the known;
$$ N= \dfrac{\dfrac{(0.20)}{(10)} }{(5\times 10^{-6})} (0.10)$$
$$ N= \color{red}{\bf 400}\;\rm turn$$
