Answer
$B = 0.33~T$
Work Step by Step
We can find $\gamma$:
$K = E_0~(\gamma-1) = 2.50~MeV$
$\gamma-1 = \frac{2.50~MeV}{E_0}$
$\gamma = 1+\frac{2.50~MeV}{0.511~MeV}$
$\gamma = 5.89$
We can find $\beta$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\sqrt{1-\beta^2} = \frac{1}{\gamma}$
$1-\beta^2 = \frac{1}{\gamma^2}$
$\beta^2 = 1-\frac{1}{\gamma^2}$
$\beta = \sqrt{1-\frac{1}{\gamma^2}}$
$\beta = \sqrt{1-\frac{1}{5.89^2}}$
$\beta = 0.9855$
We can find the magnetic field $B$:
$r = \frac{\gamma ~m~v}{\vert q \vert B}$
$B = \frac{\gamma ~m~v}{\vert q \vert r}$
$B = \frac{(5.89)~(9.109\times 10^{-31}~kg)~(0.9855)(3.0\times 10^8~m/s)}{(1.6\times 10^{-19}~C)(0.030~m)}$
$B = 0.33~T$
