For a BJT determine unity bandwidth gain and emitter base capacitance
We know that : 

$$h_{fe}= \frac{β_o}{1+s(Cπ+Cµ)r_π} = \frac{β_o}{\sqrt{1 +(Cπ+Cµ)^2\:{r_π}^2 }}$$
Additional we know thta: 
$$β_o = g_m r_{\pi}$$ 

And if we assume that this 50MHz is \$1/\omega_B\$ we have 

$$50\;\textrm{MHz} = \frac{1}{2\pi(Cπ+Cµ)r_π}$$ 

And from there we can solve for \$(Cπ+Cµ)r_π = 3.183 \cdot10^{-9}\$

So, we can solve for \$r_π\$ if \$g_m =40*I_C = 0.04 \:S \$

$$r_π = \frac{h_{fe}\sqrt{1 +\left((Cπ+Cµ)r_π\right)^2 } }{g_m} =\frac{10}{0.04} = 250\Omega$$

And if we know all of this we can solve for \$(Cπ+Cµ)\$ and \$\textrm{f}_T\$

$$(Cπ+Cµ) = \frac{(Cπ+Cµ)r_π }{r_π} = 12.72\textrm{pF} $$

$$\textrm{f}_T = \frac{g_m}{2\pi(Cπ+Cµ) }= 500\:\textrm{MHz}$$