An engine operates on the reversible cycle as shown in the figure. The work output from the engine (in kJ/cycle) is ______ (correct to two decimal places).

The correct answer is 62.5.

For any thermodynamic cycle represented on a pressure-volume ($p-V$) diagram, the net work output per cycle is equal to the area enclosed by the path of the cycle.

By analyzing the provided $p-V$ diagram, we can observe that the cycle forms a right-angled triangle. The coordinates and limits visible in the image are:
- Pressure ($p$) is plotted on the vertical axis in kilopascals ($\text{kPa}$), with values ranging from $400\text{ kPa}$ to $650\text{ kPa}$.
- Volume ($V$) is plotted on the horizontal axis in cubic meters (${\text{m}}^3$), with values ranging from $2{\text{ m}}^3$ to $2.5{\text{ m}}^3$.

We can determine the base and height of the triangular region representing the cycle:
- Base ($\Delta V$) = $2.5{\text{ m}}^3-2{\text{ m}}^3=0.5{\text{ m}}^3$
- Height ($\Delta p$) = $650\text{ kPa}-400\text{ kPa}=250\text{ kPa}$

The work output is calculated using the formula for the area of a triangle:
$\text{Work output}=\frac{1}{2}\times \text{Base}\times \text{Height}$

Substituting the values into the formula:
$\text{Work output}=\frac{1}{2}\times 0.5{\text{ m}}^3\times 250\text{ kPa}$

Since the product of pressure in $\text{kPa}$ and volume in ${\text{m}}^3$ gives energy in kilojoules ($\text{kJ}$):
$\text{Work output}=0.25\times 250=62.5\text{ kJ/cycle}$