Since the field originally has length 20 m and width 5 m, then its area is $20 \times 5=100 \mathrm{~m}^{2}$. The new length of the field is $20+10=30 \mathrm{~m}$, so the new area is $30 \times 5=150 \mathrm{~m}^{2}$. The increase in area is $150-100=50 \mathrm{~m}^{2}$. (Alternatively, we could note that since the length increases by 10 m and the width stays constant at 5 m, then the increase in area is $10 \times 5=50 \mathrm{~m}^{2}$.)