When an expansion of a system does work on its surroundings, is the sign for work negative?
It means the change in internal energy is negative, so you are on the right track. Because the system is doing work on the surroundings, that $288\ \mathrm{J}$ is a loss for the system and a gain for the surroundings.

One way to intuitively work your way through this is to think about what the signs on the terms must be in order for the problem to make sense. 

You know the final volume is greater than the initial (it's an expansion of the system into the surroundings), so $\Delta\mathrm{V}$ has to  be positive. You know that $w = -\mathrm{p_{ext}}\Delta\mathrm{V}$ is negative, because $\mathrm{p_{ext}}$ is positive. Thus, the sign of $\Delta\mathrm{E}$ is negative.

You want to solve

$$\Delta\mathrm{E} = -\mathrm{p_{ext}}\Delta\mathrm{V}$$

for $\Delta\mathrm{V}$ and add that to your initial volume of $0.25\ \mathrm{L}$.

Solving for $\Delta\mathrm{V}$:

$$-288\,\mathrm{J} = -2\ \mathrm{atm}\times\Delta\mathrm{V}$$

$$\Delta\mathrm{V} = 144\ {\mathrm{J\over atm}}$$

and then converting units

$$\Delta\mathrm{V} = 
\left(144\ {\mathrm{J\over atm}}\right)\cdot
\left({\mathrm{L\cdot atm}\over 101.325\ \mathrm{J}}\right) = 
1.42\ \mathrm{L}$$

we now solve for the final volume as (spoiler alert): 


  $$\mathrm{V_{final}} = 0.25\ \mathrm{L} + 1.42\ \mathrm{L} = 1.67\ \mathrm{L}$$