Var[2X + 3Y] = 2² Var[X] + 2 Cov[X, Y] + 3² Var[Y]
but since X and Y are given to be independent, the covariance term vanishes and you're left with
Var[2X + 3Y] = 4 Var[X] + 9 Var[Y]
X follows an exponential distribution with parameter <em>λ</em> = 1/6, so its mean is 1/<em>λ</em> = 6 and its variance is 1/<em>λ</em>² = 36.
Y is uniformly distributed over [<em>a</em>, <em>b</em>] = [4, 10], so its mean is (<em>a</em> + <em>b</em>)/2 = 7 and its variance is (<em>b</em> - <em>a</em>)²/12 = 3.
So you have
Var[2X + 3Y] = 4 × 36 + 9 × 3 = 171