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prepare 1M 100 ml H2SO4. 98% H2SO4, density 1.84

here

First of all density with no units is meaningless and makes the question impossible to solve.

Secondly the percentage without the type is meaningless.
Although can easily be find out in such question but, you should be specify that.

But to give idea about the solution lets assume
d(H2SO4) = 1.84 kg/L and 98% means
100 unit volume of stock solution contains
98 unit volume of H2SO4

so problem asks us to find the amount (mL) of H2SO4 (stock) to prepare 100 mL of 1M soluiton

x mL stock(H2SO4) =
(1mol H2SO4 / 1L solution) *
(1L solution / 1000mL solution) * (100mL solution) *
(98.04g H2SO4 / 1mol H2SO4) *
(1kg H2SO4 / 1000 g H2SO4) *
(1L H2SO4 / 1.84kg H2SO4) *
(1000mL H2SO4 / 1L H2SO4) *
(100ml stockH2SO4 / 98mL H2SO4)

----
x mL H2SO4 = 5.437 mL

to prepare 100 mL, 1M H2SO4 from stock (98% [v/v]; 1.84 [kg/L]; 98.04 [g/mol])

to 100 mL flat bottom flask add some amount of water (not much),
and pour 5.45 mL H2SO4 stock solution on it.
Dilute the solution to 100 mL

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