This looks like homework, so here are some pointers :-
Firstly, you need to find the probability density function P(x) for the signal, given that :-
\$\int\limits_{-a}^{+a} P(x)\,\mathrm dx=1\$
The variance is given by :-
\$\int\limits_{-a}^{+a} x^2P(x)\,\mathrm dx\$
since the mean is zero.
The rms is the square-root of this by definition (as Olin says), so you have your answer in terms of a.
So my question is if we have a pulse wave does it make sense anymore
to talk about its RMS value?
Yes, of course it does; the rms value of the pulse wave is the effective DC voltage across a resistor that gives the same average power.
Recall that the instantaneous power associated with a resistor is
$$p_R(t) = \dfrac{v^2_R(t)}{R} $$
The average power, over a period \$T\$, is then
$$p_{avg} = \dfrac{1}{T} \int_0^Tp_R(t)\,dt =\dfrac{1}{T} \int_0^T\dfrac{v^2_R(t)}{R}\,dt$$
Thus, the equivalent DC voltage that produces the same average power is
$$V_{eq} = \sqrt{p_{avg}\cdot R} = \sqrt{\dfrac{1}{T} \int_0^Tv^2_R(t)\,dt}$$
But, that last term is precisely the root of the mean of the square (rms) value of \$v_R(t)\$.
So, yes, it makes sense to talk about the rms value of a pulse waveform or any other voltage or current waveform for that matter.
Best Answer
The heating effect refers to the power that would be dissipated if that current flowed in a pure resistor. Such a resistor would convert all of its dissipated power into heat. The RMS value of a current is equal to a DC current that produces the same amount of heat in that resistor.