The forces on the beaker+water system are
gravity, acting downward (100g)
the force you exert to counterbalance the buoyancy of the pencil (a volume of 2.5cm^3, a density of 0.5g/cm^3, so effective weight of 1.25g)
balancing those, the upward reaction force from the weighing scale, which determines what weight the scale reads.
So I think the scale will measure
(I am adopting the convenient approximation, already present in the question itself, that the density of water is exactly 1g/cm^3, which isn't exactly right but is very close.)
But that's not quite right because
the water level will rise a bit when the pencil goes in, so the volume of water displaced by the pencil is slightly larger than the figure above. (Thanks to Daniel Mathias in comments for noticing this; I missed it.) Let's be more precise. The beaker contained 100cm^3 of water at a height of 5cm so its base area is 20cm^2; now a column of base-area 0.5cm^2 contains pencil rather than water, so the area has decreased by a ratio of 39/40 and the height has therefore increased by 40/39. So the depth is now 200/39cm or about 5.13cm, so instead of 1.25g as above we need 40/39*1.25g=50/39g ~= 1.28g. So the scale, if accurate enough, will measure 101.28g instead of 101.25g.