After fifth collision, node will choose a value of K at random from {0, 1, 2,..., 25 - 1} i.e. 0 to 31.
So P(K=4) = $\frac{1}{32}$
Now, after choosing K=4, node will wait for K*512 bit times. Bit time for 10 Mbps ethernet is 0.1microsec. Therefore total waiting time for node = 4 * 512 * 0.1 microsec = 204.8 microsec.