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 λ = 1/6, so its mean is 1/λ = 6 and its variance is 1/λ² = 36.
Y is uniformly distributed over [a, b] = [4, 10], so its mean is (a + b)/2 = 7 and its variance is (b - a)²/12 = 3.
So you have
Var[2X + 3Y] = 4 × 36 + 9 × 3 = 171
