Answer
$917\;\rm g$
Work Step by Step
So, we need to draw the force diagrams of the hamster and the inclined (wedge-shaped block), as we see below. Recall that the friction force between the hamster and the inclined block is zero since there is a lubricating oil having of $\mu_s=\mu_k=0$.
We know that the scale reading when the hamster is at rest at the top of the wedge, and before adding the lubricating oil, is the sum of the weight of the two of the hamster and the wedge.
$$\dfrac{F_{n0}}{g}=m_{H}+m_W=200+800=\bf 1000\;\rm g$$
We also know that the normal force exerted on the wedge-shaped block is the reading of the scale since it is the reaction force exerted by the scale on the system.
So that to find the reading of the scale we need to find the normal force exerted on the inclined.
Note that the friction force between the scale and the wedge had to be toward the right since the reaction force exerted by the hamster has a negative $x$-components, as you see below on the right side of the figures.
Applying Newton's second law on the hamster;
$$\sum F_{yH}=F_{\rm (W\;on\;H)}-m_Hg\cos\theta=m_Ha_{xH}=m_H(0)=0$$
Thus,
$$F_{\rm (W\;on\;H)}=m_Hg\cos\theta\tag 1$$
Applying Newton's second law on the wedge;
$$\sum F_{xW}=f_s-F_{\rm (W\;on\;H)}\sin\theta =m_Wa_{xW}=m_W(0)=0$$
Thus,
$$f_s=F_{\rm (W\;on\;H)}\sin\theta \tag 2$$
$$\sum F_{yW}=F_n-m_Wg-F_{\rm (W\;on\;H)}\cos\theta=m_Wa_{yW}=m_W(0)=0$$
Thus,
$$F_n=m_Wg+F_{\rm (W\;on\;H)}\cos\theta $$
Plugging from (1);
$$F_n=m_Wg+m_Hg\cos\theta\cos\theta $$
$$F_n=m_Wg+m_Hg\cos^2\theta $$
Plugging the known;
$$F_n=(0.8\times9.8)+(0.2\times9.8 \cos^240^\circ) $$
$$F_n=\bf 8.99\;\rm N$$
Therefore the reading of the scale in grams is given by
$$\text{Scale reading}=\dfrac{F_n}{g}\times \dfrac{1000\;\rm g}{1\;\rm kg}$$
Plugging the known;
$$\text{Scale reading}=\dfrac{8.99}{9.8}\times \dfrac{1000\;\rm g}{1\;\rm kg}$$
$$\text{Scale reading}=\color{red}{\bf 917}\;\rm g$$
