There are two event chains that lead to the second ball selected being red.

(select W, select R)

(select R, select R)

let's look at probabilities of the first ball selection.

$Pr[W]=\dfrac 6 7$

$Pr[R]=\dfrac 1 7$

Now

$Pr[R|W]=\dfrac 1 6$ as the white urn contains (5 white, 1 red) after having 1 white ball removed.

$Pr[R|R] = \dfrac 6 7$ as it has been untouched

So our total probability

$Pr[R \mbox{ is second selection}]=Pr[R|W]Pr[W] + Pr[R|R]Pr[R] = \dfrac 1 6 \dfrac 6 7 + \dfrac 6 7 \dfrac 1 7 = \dfrac {13}{49}$