I'm trying to shown that,
$$E\left[\left(\int_0^t sdW_s\right)^2\lvert W_t\right]=\frac{W_t^2t^2}{4}+\frac{t^3}{12}$$
I know that first term is the conditional expectation squared, remaining to show that the second term is the conditional variance. I will be very glad if you could give me any hint.
As it was mentioned in a comment $$d(tW_t) = tdW_t + W_tdt,$$ so $$\int_0^t sdW_s = tW_t-\int_0^tW_sds.$$ Moreover, it's well-known that $$\mathbb E[f((W_s)_{s\leq t})\mid W_t] = \mathbb E[f((Y_s^t+\frac stx)_{s\leq t})]_{x=W_t},$$ where $(Y_s^t)_{s\leq t}$ is a Brownian bridge. This just means that $(W_s)_{s\leq t}\mid W_t$ is a Brownian bridge starting at $0$ and ending at $W_t$.
Thus, $$\mathbb E\Big[\int_0^t W_sds\mid W_t\Big] = \mathbb E\Big[\int_0^tY_s^tds\Big] + \frac 1t\mathbb E\Big[\int_0^tsxds\Big]_{x=W_t} = 0 + \frac{tW_t}{2} = \frac{tW_t}2$$ and $$\mathbb E\Big[\Big(\int_0^t W_sds\Big)^2\mid W_t\Big] = \mathbb E\Big[\Big(\int_0^t Y_s^tds\Big)^2\Big] + \frac1{t^2}\mathbb E\Big[\Big(\int_0^t sxds\Big)^2\Big]_{x=W_t} + \frac2t\mathbb E\Big[\Big(\int_0^tY_s^tds\Big)\Big(\int_0^tsxds\Big)\Big]_{x=W_t} = \int_0^t\int_0^t\mathbb E[Y_s^tY_r^t]dsdr+ \frac{t^2W_t^2}{4},$$ which, since the covariance of the Brownian bridge is $s\wedge r - \frac{sr}t$, is equal to $$ \frac{t^2W_t^2}{4} + 2\int_0^t\int_0^sr(1-\frac st)drds = \frac{t^2W_t^2}{4} + \int_0^ts^2(1-\frac st)ds = \frac{t^2W_t^2}{4} + \frac{t^3}{12}.$$ Finally, $$\mathbb E\Big[\Big(\int_0^tsdW_s\Big)^2\mid W_t\Big] = t^2W_t^2 - 2tW_t\mathbb E\Big[\int_0^tW_sds\mid W_t\Big] + \mathbb E\Big[\Big(\int_0^tW_sds\Big)^2\mid W_t\Big] = t^2W_t^2 - t^2W_t^2 + \frac{t^2W_t^2}{4} + \frac{t^3}{12} = \frac{t^2W_t^2}{4} + \frac{t^3}{12}.$$