Answer: B

Solution:

Multiples of $3$ or $4$ up to $2001:$ $\left\lfloor\tfrac{2001}{3}\right\rfloor$ $+\left\lfloor\tfrac{2001}{4}\right\rfloor$ $-\left\lfloor\tfrac{2001}{12}\right\rfloor$ $=667+500-166$ $=1001.$

Among these, remove the multiples of $5:$ multiples of $15$ ($133$) and of $20$ ($100$), re-adding multiples of $60$ ($33$): $133+100-33=200.$

So the count is $1001-200=801.$ Thus, the correct answer is B.