Answer
$\sqrt{\frac{1}{\alpha}}$
Work Step by Step
We know that the tangential acceleration is given as
$a_t=R\alpha$
and the centripetal acceleration is given as
$a_{cp}=R\omega^2$
$a_{cp}=R(\alpha t)^2$
$a_{cp}=R\alpha^2t^2$
Given that $a_t=a_{cp}$
$\implies R\alpha=R\alpha^2t^2$
$\implies I=\alpha t^2$
This can be rearranged as:
$t^2=\frac{1}{\alpha}$
$\implies t=\sqrt{\frac{1}{\alpha}}$