Given P(getting Head) = p which implies P(not getting head) = 1-p
Also given that number of tosses required are even for the head to appear for first time.
So, getting first head on $2^{nd} $ toss = $(1-p)*(p)$ OR
getting first head on $4^{th}$ toss = $(1-p)^{3}*(p)$ OR ....
Thererfore we obtain an infinite GP series i.e
$(1-p)*(p)$ + $(1-p)^{3}*(p)$ + $(1-p)^{5}*(p)$+.............upto infinity = $2/5$
Solving above series we get $ p = 1/3$