Let $X$ be a r.v. denoting the number of repetitions of the experiment. Then, $X$ can have values $0,1,2,...$
Probability that the experiment is repeated $0$ times (i.e., not repeated at all) is given by $p(nr)=\frac{1}{10}+\frac{3}{10}+\frac{3}{10}=\frac{7}{10}$ This is obtained by summing up the three possibilities of not getting 2 white balls - either both red, or 1st red, 2nd white or 1st white, 2nd red.
That means, the whole exercise would be repeated with a probability of $\frac{3}{10}$.
Expected number of repeats is given by $E(X)=(0*\frac{7}{10})+(1*\frac{3}{10})+(2*(\frac{3}{10})^2)+...$
That is, $E(X)=\sum_{x=1}^{\infty}x(\frac{3}{10})^x$
This is an infinite G.P. with sum given by $\frac{\frac{3}{10}}{1-\frac{3}{10}}=\frac{3}{7}$
The expected number of repeats would be 1.