Answer
$B=21.1{\mu}T$
Work Step by Step
As we know that,
$B =\frac{mv}{qR}$
We know the value of mass and charge of an electron. We also know $v$ and $R$ from the data in the question. We plug in these values in the formula to obtain:
$B=\frac{9.109\times 10^{-31}(1.30\times 10^6)}{1.6\times 10^{-19}(0.350)}$
$B=2.11\times 10^{-5}$
$B=21.1\times 10^{-6}$
$B=21.1{\mu}T$
