Answer
$E=1.87\times 10^6\frac{V}{m}$
It is perpendicular to velocity and magnetic field and in opposite direction to magnetic force.
Work Step by Step
In the given scenario:
$F_E=F_B$ where $F_E$ and $F_B$ represent electric and magnetic force respectively.
$qE=qvB$
This can be rearranged as:
$E=vB$.......eq(1)
We also know that in the given scenario, magnetic force is equal to the centripetal force
$qvB=\frac{mv^2}{r}$
This simplifies to
$v=\frac{qBr}{m}$
Thus equation(1) becomes
$E=(\frac{qBr}{m})B$
$E=\frac{qB^2r}{m}$
We plug in the known values to obtain:
$E=\frac{1.6\times 10^{-19}(0.566)^2\times 6.1\times 10^{-2}}{1.67\times 10^{-27}}=1.87\times 10^6\frac{V}{m}$
It is perpendicular to velocity and magnetic field and in opposite direction to magnetic force.