Answer
The radial speed of the galaxy relative to the Earth is $~~v = 0.062~c$
Work Step by Step
We can find $\beta$:
$\lambda = \lambda_0~\sqrt{\frac{1+\beta}{1-\beta}}$
$\frac{\lambda}{\lambda_0} = \sqrt{\frac{1+\beta}{1-\beta}}$
$\frac{462~nm}{434~nm} = \sqrt{\frac{1+\beta}{1-\beta}}$
$1.064516 = \sqrt{\frac{1+\beta}{1-\beta}}$
$1.1332 = \frac{1+\beta}{1-\beta}$
$1.1332-1.1332~\beta = 1+\beta$
$2.1332~\beta = 0.1332$
$\beta = \frac{0.1332}{2.1332}$
$\beta = 0.062$
The radial speed of the galaxy relative to the Earth is $~~v = 0.062~c$
