Finding potential using spherical harmonics

I have been trying to solve the following question:

The potential on the surface of a sphere is given by
$mathbf {V = V_{0} sin^2theta sin2phi,;}$
find the potential outside the sphere

I am trying to solve it by separation of variable in spherical coordinates by using the following formula for potential outside the sphere,

$$V=sum_{l=0}^inftyfrac{B_{lm}}{r^{l+1}} {Y_l}^m (theta,phi)$$

Now the potential on the surface of the sphere is given, so we can use that for r=R as,

$$tag{1}V_{0}sin^2thetasin 2phi=sum_{l=0}^inftyfrac{B_{lm}}{R^{l+1}} {Y_l}^m (theta,phi)$$

Next for the value of $B_l$ I multiply both side with ${Y^*_l}^m$ and integrate. RHS becomes $frac{B_{lm}}{r^{l+1}}$ while LHS becomes interesting. I note that $sin^2theta$ $sin 2phi$ can be converted into $Y_2^2$ with some Constant factor. $Y_2^2$ is given as follows: $$ Y_2^2= A sin^2theta e^{imphi}$$
So my problem is, can I some how convert this into $Y_2^2$ so that it simply gives me the left hand side of equation ${(1)}?;$ I see that $sin2phi$ is the imaginary part of $e^{imphi}$ with $m=2$. Please guide me through this.