Calculate the mass of Mg needed to produce 5 kg of vanadium
You answer is correct.

It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.

First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:

$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$

Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:

$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$

Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:

$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$