Answer
The diameter would have to be $1.73~D$.
Work Step by Step
We can write an expression for the Young's Modulus of the wire for the first condition.
$Y = \frac{F_1/A_1}{\Delta~L/L_0}$
We can write an expression for the Young's Modulus of the wire for the second condition.
$Y = \frac{F_2/A_2}{\Delta~L/L_0}$
The Young's Modulus is the same in both cases since the material is the same, so we can equate the two expressions.
$\frac{F_1/A_1}{\Delta~L/L_0}=\frac{F_2/A_2}{\Delta~L/L_0}$
$A_2 = \frac{F_2~A_1}{F_1}$
$A_2 = \frac{(3~F_1)~A_1}{F_1}$
$A_2 = 3~A_1$
The area of the wire in the second condition must be three times larger. Therefore, the radius (and thus the diameter) needs to increase by a factor of $\sqrt{3}$, (since $Area = \pi~r^2$). Note that $\sqrt{3} = 1.73$
Therefore, the diameter would have to be $1.73~D$.
