$V_{opt}$ is clearly $60$. You can go for brute force or by normal intuition you can get it.
Now solving for $V_{greedy}$.
$$\begin{array}{|c|c|c|c|}\hline \textbf{Item name} & \textbf{Weight (in Kgs) }& \textbf{Value (in Rupees)} & \textbf{Value/Weight} \\\hline \text{1} & \text{10} & \text{60} & \text{6} \\\hline \text{2} & \text{7} & \text{28} & \text{4} \\\hline \text{3} & \text{4} & \text{20} & \text{5} \\\hline \text{4} & \text{2} & \text{24} & \text{12} \\\hline \end{array}$$
Sort them in descending order of Value/Weight as per the question.
$$\begin{array}{|c|c|c|c|}\hline \textbf{Item name} & \textbf{Weight (in Kgs) }& \textbf{Value (in Rupees)} & \textbf{Value/Weight} \\\hline \text{4} & \text{2} & \text{24} & \text{12} \\\hline \text{1} & \text{10} & \text{60} & \text{6} \\\hline \text{3} & \text{4} & \text{20} & \text{5} \\\hline \text{2} & \text{7} & \text{28} & \text{4} \\\hline \end{array}$$
Now start picking items.(Note: You cannot take a fraction of the given weight as per the question). Max weight size is given as $11$(Inclusive).
- Item $4$ is picked. Weight remaining = $11-2 = 9$kg.
- Item $1$ cannot be picked as $10$kg $>9$kg.
- Item $3$ can be picked as $4$kg $<$ $9$kg. Weight Remaining = $9-4 = 5$kg
- Item $2$ cannot be picked as $7$kg $>5$kg.
So, item $4$ and Item $3$ are picked. Their values are $24$ and $20$ respectively.
$\implies V_{greedy} = 24+20 = 44.$
$V_{optimal} - V_{greedy}= 60 - 44 = 16.$