Let the first number be 'x' and the second number be 'y'

According to the question 

Sum of twice the first number and thrice the second number is 141 = 2x+3y=141  ---Eq(1) 

Sum of thrice the first number and twiice the second number is 174 = 3x+2y=174 ---Eq(2)

From Eq(1) 

$$\begin{gathered} = 2x+3y=241 \\ = x=58-\frac{2}{3}y ---eq\left(3\right) \\ = 2\left(58-\frac{2}{3}y\right)+3y=141 \\ = y=15 \\ Substitue \text{'}y\text{'} value in Eq\left(3\right) we get, \\ = x=58-\frac{2}{3}\times \left(15\right) \\ =x=48 \\ \therefore (x,y)=(48,15) \end{gathered}$$

The largest number is x=48

Hence, Option C is ccorrect.